Found problems: 1065
Let $ABC$ be an acute triangle and $AD$ a height. The angle bissector of $\angle DAC$ intersects $DC$ at $E$. Let $F$ be a point on $AE$ such that $BF$ is perpendicular to $AE$. If $\angle BAE=45º$, find $\angle BFC$.
Two circles intersect at points $A$ and $B$. A common tangent touches the first circle at point $C$ and the second at point $D$. Let $\angle CBD > \angle CAD$. Let the line $CB$ intersect the second circle again at point $E$. Prove that $AD$ bisects the angle $\angle CAE$.
(P Kozhevnikov)
Let $ABC$ be an acute-angled triangle inscribed in circle $(O)$. Let $G$ be a point on the small arc $AC$ of $(O)$ and $(K)$ be a circle passing through $A$ and $G$. Bisector of $\angle BAC$ cuts $(K)$ again at $P$. The point $E$ is chosen on $(K)$ such that $AE$ is parallel to $BC$. The line $PK$ meets the perpendicular bisector of $BC$ at $F$. Prove that $\angle EGF = 90^o$.
The diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$ intersect at point $ M$. The bisector of $ \angle ACD$ meets the ray $ BA$ at $ K$. Given that $ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$, prove that $ \angle BKC \equal{} \angle CDB$.
Construct triangle $ABC$, given the altitude and the angle bisector both from $A$, if it is known for the sides of the triangle $ABC$ that $2BC = AB + AC$.
(Alexey Karlyuchenko)
Let $P$ be a given point inside a convex quadrilateral $O_1O_2O_3O_4$. For each $i = 1,2,3,4$, consider the lines $l$ that pass through $P$ and meet the rays $O_iO_{i-1}$ and $O_iO_{i+1}$ (where $O_0 = O_4$ and $O_5 = O_1$) at distinct points $A_i(l)$ and $B_i(l)$, respectively. Denote $f_i(l) = PA_i(l) \cdot PB_i(l)$. Among all such lines $l$, let $l_i$ be the one that minimizes $f_i$. Show that if $l_1 = l_3$ and $l_2 = l_4$, then the quadrilateral $O_1O_2O_3O_4$ is a parallelogram.
Angle bisectors $AD$ and $BE$ are drawn in triangle $ABC$. It turned out that $DE$ is the bisector of triangle $ADC$. Find the angle $BAC$.
Let $ABCD$ be a parallelogram. Let $M$ be a point on the side $AB$ and $N$ be a point on the side $BC$ such that the segments $AM$ and $CN$ have equal lengths and are non-zero. The lines $AN$ and $CM$ meet at $Q$.
Prove that the line $DQ$ is the bisector of the angle $\measuredangle ADC$.
[i]Alternative formulation.[/i] Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$.
Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$.
Let $ABC$ be a triangle such that $|AB|=8$ and $|AC|=2|BC|$. What is the largest value of altitude from side $[AB]$?
$
\textbf{(A)}\ 3\sqrt 2
\qquad\textbf{(B)}\ 3\sqrt 3
\qquad\textbf{(C)}\ 5
\qquad\textbf{(D)}\ \dfrac {16}3
\qquad\textbf{(E)}\ 6
$
Let $ABC$ be an acute triangle and $A_{1}, B_{1}, C_{1}$ be the touchpoints of the excircles with the segments $BC, CA, AB$ respectively. Let $O_{A}, O_{B}, O_{C}$ be the circumcenters of $\triangle AB_{1}C_{1}, \triangle BC_{1}A_{1}, \triangle CA_{1}B_{1}$ respectively. Prove that the lines through $O_{A}, O_{B}, O_{C}$ respectively parallel to the internal angle bisectors of $\angle A,\angle B, \angle C$ are concurrent.
Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \neq CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \neq C$ on the line $CL$, we have $\angle XAC \neq \angle XBC$. Also show that for $Y \neq C$ on the line $CH$ we have $\angle YAC \neq \angle YBC$.
[i]Bulgaria[/i]
Let $ABCD$ be a cyclic quadrilateral.$E$ is the midpoint of $(AC)$ and $F$ is the midpoint of $(BD)$ {$G$}=$AB\cap CD$ and {$H$}=$AD\cap BC$.
a)Prove that the intersections of the angle bisector of $\angle{AHB}$ and the sides $AB$ and $CD$ and the intersections of the angle bisector of$\angle{AGD}$ with $BC$ and $AD$ are the verticles of a rhombus
b)Prove that the center of this rhombus lies on $EF$
Given an acute angle $APX$ in the plane, construct a square $ABCD$ such that $P$ lies on the side $BC$ and ray $PX$ meets $CD$ in a point $Q$ such that $AP$ bisects the angle $BAQ$.
$O$ is a circumcircle of non-isosceles triangle $ABC$ and the angle bisector of $A$ meets $BC$ at $D$. If the line perpendicular to $BC$ passing through $D$ meets $AO$ at $E$, show that $ADE$ is an isosceles triangle.
$ABC$ is an acute-angled triangle. The tangents to the circumcircle at $A$ and $C$ meet the tangent at $B$ at $M$ and $N$. The altitude from $B$ meets $AC$ at $P$. Show that $BP$ bisects the angle $MPN$
Let $a,b,c$ be the length sides of a given triangle and $x,y,z$ be the sides length of bisectrices, respectively. Prove the following inequality $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Draw a line passing through a point $M$ on the angle bisector of the angle $\angle AOB$, that intersects $OA$ and $OB$ at points $K$ and $L$ respectively. Prove that the valus of the sum $\frac{1}{|OK|}+\frac{1}{|OL|}$ does not depend on the choice of the straight line passing through $M$, i.e. is defined by the size of the angle AOB and the selection of the point $M$ only.
Suppose that $ B_1$ is the midpoint of the arc $ AC$, containing $ B$, in the circumcircle of $ \triangle ABC$, and let $ I_b$ be the $ B$-excircle's center. Assume that the external angle bisector of $ \angle ABC$ intersects $ AC$ at $ B_2$. Prove that $ B_2I$ is perpendicular to $ B_1I_B$, where $ I$ is the incenter of $ \triangle ABC$.
Circles $\alpha$ and $\beta$ of the same radius intersect in two points, one of which is $P$. Denote by $A$ and $B$, respectively, the points diametrically opposite to $P$ on each of $\alpha$ and $\beta$ . A third circle of the same radius passes through $P$ and intersects $\alpha$ and $\beta$ in the points $X$ and $Y$ , respectively. Show that the line $XY$ is parallel to the line $AB$.
Given a triangle $ ABC$, let $ r$ be the external bisector of $ \angle ABC$. $ P$ and $ Q$ are the feet of the perpendiculars from $ A$ and $ C$ to $ r$. If $ CP \cap BA \equal{} M$ and $ AQ \cap BC\equal{}N$, show that $ MN$, $ r$ and $ AC$ concur.
In triangle $ABC$, the midpoints of sides $AB$ and $AC$ are $D$ and $E$, respectively. Prove that the bisectors of the angles $BDE$ and $CED$ intersect at the side $BC$ if the length of side $BC$ is the arithmetic mean of the lengths of sides $AB$ and $AC$.
Point $M$ is the midpoint of base $AD$ of an isosceles trapezoid $ABCD$ with circumcircle $\omega$. The angle bisector of $ABD$ intersects $\omega$ at $K$. Line $CM$ meets $\omega$ again at $N$. From point $B$, tangents $BP, BQ$ are drawn to $(KMN)$. Prove that $BK, MN, PQ$ are concurrent.
[i]A. Kuznetsov[/i]
In $\triangle ABC$, let $AD$ be the angle bisector of $\angle BAC$, with $D$ on $BC$. The perpendicular from $B$ to $AD$ intersects the circumcircle of $\triangle ABD$ at $B$ and $E$. Prove that $E$, $A$ and the circumcenter $O$ of $\triangle ABC$ are collinear.
In triangle $ABC$, the angle bisectors are $BE$ and $CF$ (where $E, F$ are on the sides of the triangle), and their intersection point is $I$. Point $N$ lies on the circumcircle of $AEF$, and the angle $\angle IAN$ is right. The circumcircle of $AEF$ meets the line $NI$ a second time at the point $L$. Show that the circumcenter of $AIL$ lies on line $BC$.
In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$.
[i]Proposed by Ray Li[/i]